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Wolfram Alpha symbolic integer derivative to this question is wrong.

How can i obtain it by Mathematica? Thanks

 D[1/(-x + x*Exp[I*Pi*x]), {x, n}]
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  • $\begingroup$ @Daniel Lichtblau,Maybe you open this question? $\endgroup$ Commented Feb 8, 2017 at 7:28
  • $\begingroup$ In V13, the OP's code D[1/(-x + x*Exp[I*Pi*x]), {x, n}] now works. $\endgroup$
    – Michael E2
    Commented Dec 25, 2021 at 18:25

1 Answer 1

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From Maple:

$\sum _{m=0}^n \binom{n}{m} (-m)_m x^{-1-m} \frac{\partial ^{m-n}}{\partial x^{m-n}}\frac{1}{\exp (i \pi x)-1}$

Finding n-th derivative:

$f(x)=\frac{1}{-1+e^{i \pi x}}$

f[x_] := 1/(Exp[Pi*I*x] - 1)
func = SeriesCoefficient[f[x], {x, 0, m}]
func = FullSimplify[func, {m \[Element] Integers, m > 0}]

enter image description here

$$f(x)=\sum _{m=-1}^{\infty } \text{SeriesCoefficient}[f(x),\{x,0,m\}] x^m$$ $$\frac{\partial ^nf(x)}{\partial x^n}=\sum _{m=-1}^{\infty } \text{SeriesCoefficient}[f(x),\{x,0,m\}] \frac{\partial ^nx^m}{\partial x^n}$$ $$\frac{\partial ^nf(x)}{\partial x^n}=\sum _{m=-1}^{\infty } \text{SeriesCoefficient}[f(x),\{x,0,m\}] \binom{m}{m-n} n! x^{m-n}$$

Sum[func*Binomial[m, m - n]*n!*x^(m - n), {m, -1, Infinity}]

 

MMA can't find sum :(. Changing index m to m-1 in sum to be m=0:

$$\frac{\partial ^n\frac{1}{(-1+e^{i \pi x})}}{\partial x^n}=\sum _{m=0}^{\infty } \frac{\left((i \pi )^{m-1} B_m\right) \binom{m-1}{m-n-1} n! x^{m-n-1}}{\Gamma (1+m)}$$

Substituting to first sum.

$$\frac{\partial ^n}{\partial x^n}(\frac{1}{-x+x \exp (i \pi x)})=\sum _{m=0}^n \binom{n}{m} (-m)_m x^{-1-m} \sum _{j=0}^{\infty } \frac{\left((i \pi )^{j-1} B_j\right) \binom{j-1}{j-n+m-1} (n-m)! x^{j-n+m-1}}{\Gamma (1+j)}$$

Check numerics:

n = 2;
N[D[1/(-x + x*Exp[I*Pi*x]), {x, n}] /. x -> 1, 50]
(*-1.0000000000000000000000000000000000000000000000000 - 
1.5707963267948966192313216916397514420985846996876 I*)

N[Sum[Binomial[n, m]*Pochhammer[-m, m]*x^(-1 - m)*
Sum[((I \[Pi])^(j - 1) BernoulliB[j])/Gamma[1 + j]*
Binomial[j - 1, j - n + m - 1]*(n - m)!*x^(j - n + m - 1), {j, 
0, 1000}], {m, 0, n}] /. x -> 1, 50]
(*-1.0000000000000000000000000000000000000000000000000 - 
1.5707963267948966192313216916397514420985846996876 I*)

It seems that the result is correct.

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  • $\begingroup$ And how you know the n-th derivative of 1/(-1+e^(i pi x)) in mathematica since wolfram alpha dont answers that? $\endgroup$
    – Kevin67
    Commented Jan 11, 2017 at 15:36
  • $\begingroup$ Maybe send your question here: math.stackexchange.com .There you will get the answer, if it is possible. $\endgroup$ Commented Jan 11, 2017 at 15:55
  • $\begingroup$ $Kevin67.I'm updated the answer. :) $\endgroup$ Commented Feb 7, 2017 at 12:15

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